64 drops of water are combined to form one big drop. Surface energy of big drop is how many times of any small drop.

Options

A. 16
B. 8
C. 4
D. None

Explanation

To solve the problem of how many times the surface energy of a big drop is compared to that of a small drop, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Volume Relationship**:
- We have 64 small drops that combine to form one big drop.
- The volume of a drop is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
- Let \( r \) be the radius of a small drop and \( R \) be the radius of the big drop.
- The total volume of the 64 small drops is:
\[ V_{\text{small}} = 64 \times \frac{4}{3} \pi r^3 \]
- The volume of the big drop is:
\[ V_{\text{big}} = \frac{4}{3} \pi R^3 \]

2. **Setting the Volumes Equal**:
- Since the volume of the big drop is equal to the total volume of the small drops, we can set the equations equal:
\[ \frac{4}{3} \pi R^3 = 64 \times \frac{4}{3} \pi r^3 \]
- We can cancel \(\frac{4}{3} \pi\) from both sides:
\[ R^3 = 64 r^3 \]

3. **Finding the Radius of the Big Drop**:
- Taking the cube root of both sides gives:
\[ R = 4r \]

4. **Calculating Surface Areas**:
- The surface area \( S \) of a drop is given by:
\[ S = 4 \pi r^2 \]
- For the small drop:
\[ S_{\text{small}} = 4 \pi r^2 \]
- For the big drop:
\[ S_{\text{big}} = 4 \pi R^2 = 4 \pi (4r)^2 = 4 \pi \times 16r^2 = 64 \pi r^2 \]

5. **Calculating Surface Energy**:
- The surface energy \( E \) is given by:
\[ E = \text{Surface Tension} \times \text{Surface Area} \]
- Let \( \gamma \) be the surface tension (which is the same for both drops).
- The surface energy for the small drop:
\[ E_{\text{small}} = \gamma \times S_{\text{small}} = \gamma \times 4 \pi r^2 \]
- The surface energy for the big drop:
\[ E_{\text{big}} = \gamma \times S_{\text{big}} = \gamma \times 64 \pi r^2 \]

6. **Finding the Ratio of Surface Energies**:
- Now, we can find the ratio of the surface energy of the big drop to that of a small drop:
\[ \frac{E_{\text{big}}}{E_{\text{small}}} = \frac{\gamma \times 64 \pi r^2}{\gamma \times 4 \pi r^2} = \frac{64}{4} = 16 \]

### Final Answer:
The surface energy of the big drop is **16 times** that of any small drop.